Thailand's Logistics Costs: Challenges and Solutions

1. การขนส่ง(Transportation)

2. เป็นการเคลื่อนย้ายสินค้าและบริการจากแหล่งผู้ผลิตหรือผู้จัดเก็บ ไปยังลูกค้าในระดับต่างๆ เชื่อมต่อโซ่อุปทาน การขนส่ง

3. เป็นกิจกรรมที่ไม่ได้เพิ่มมูลค่าให้กับสินค้าโดยตรงเป็นกิจกรรมที่ไม่ได้เพิ่มมูลค่าให้กับสินค้าโดยตรง เพิ่มมูลค่าของผลิตภัณฑ์ในด้านสถานที่(Place Value Added) สิ่งที่ต้องพิจารณาในการขนส่ง ปริมาณที่ขนส่ง เวลาที่ขนส่ง ความถี่ในการขนส่ง

4. THAILAND’S LOGISTICS COSTS PER GDP

6. วิกฤตพลังงาน ทำให้น้ำมันมีราคาสูง  ต้นทุนด้านการขนส่งเพิ่มสูงขึ้น • รายงานโลจิสติกส์ของประเทศไทยประจำปี 2553 • (สำนักงานคณะกรรมการพัฒนาการเศรษฐกิจและสังคมแห่งชาติ) • ในปี 2551 ต้นทุนโลจิสติกส์ = 15.2% ของ GDP (1.64 ล้านล้านบาท) ต้นทุนการขนส่งสินค้า = 7.2% ของ GDP (776.4 พันล้านบาท) ต้นทุนการเก็บรักษาสินค้าคงคลัง = 6.7% ของ GDP (722.5 พันล้านบาท) ต้นทุนการการบริหารจัดการ = 1.3% ของ GDP (145.1 พันล้านบาท) ต้นทุนการขนส่งสินค้าค่าน้ำมันเชื้อเพลิง จัดเส้นทางการขนส่งสินค้าที่เหมาะสม ลดค่าน้ำมันเชื้อเพลิง

7. รูปแบบการขนส่ง ทางน้ำ ทางอากาศ ทางถนน ทางท่อ ทางรถไฟ

8. S7 S6 S3 S4 Warehouse S2 S5 S8 S1 Traveling Salesman Problem (TSP) S9

9. S6 S7 S10 S3 S4 Warehouse S2 S5 S8 S1 S9 Vehicle Routing Problem (VRP)

10. Depot Customer 2 7 1 5 4 10 9 6 8 3 ปัญหาการเดินทางของพนักงานขาย Traveling Salesman Problem-TSP 0 • One “depot” node (origin and destination). • Multiple demand nodes and one vehicle. • Find a tour that visits each demand node exactly once. • Objective: Minimize the total length of the tour.

11. Depot Customer 2 7 1 5 4 10 9 6 8 3 ปัญหาการเดินทางของพนักงานขาย Traveling Salesman Problem-TSP 0 Assumptions 1. Undirected and symmetric arcs. 2. All possible arcs between nodes exist. 3. Each node is visited exactly once. 4. For transportation problems, a vehicle has unlimited capacity.

12. Mathematical Formulation as Integer Programming Subject to:

13. Select node j that is nearest node i and not in a subtour. Add node j to the tour (connected to node i) and set i = j. Yes Is there aNode not in a tour? No End TSP Heuristic Approaches Basic Nearest Neighbor Heuristic Randomly select an initial node i as a partial tour. j i 2 1 j i 3 0 8 4 5 7 6

14. TSP Basic Nearest Neighbor Heuristic 0 5 3 1 6 2 4 7 0 Total distance = 10+11+10+30+10+20+10+90 = 191

15. Compare the total distance (cost) of all tourswith a different starting point. Choose the tour with smallest total distance as the solution. TSP Modified Nearest Neighbor Heuristic For every node i as the starting point of the tour, construct the tour using the basic nearest neighbor heuristic. 50 52 i 2 1 53 55 3 54 0 8 53 4 55 5 7 51 6 56

16. Randomly select a node not in the tour. Insert that node in the edge that incurs smallest increasing distance (cost). Yes Is there aNode not in a tour? No End TSP Arbitrary Insertion Heuristic Select a starting tour with k nodes (k ≥ 1). 2 1 3 0 8 4 5 7 6

17. Heuristics for the asymmetric TSP The savings method -Develop in the context of the veicle routing problem. -Studied by Golden (1977), Golden, Bodin, Doyle & Stewart (1980) Step 1. Select any city as an origin and denote this as city 1. Step 2. Compute savings Step 3. Order the savings from largest to smallest. Step 4. Starting at the top of the savings list and moving downwards, from larger subtours by linking appropriate cities i and j . Repeat until a tour is formed.

18. The loss method -Proposed by Webb (1971) for the symmetric TSP and by Van Der Cruyssen & Rijckaert (1978) for the asymmetric TSP. Step 1. Compute a loss or opportunity cost for each city j other than those that have both an indegree and an outdegree. Step 2. Connect the city having the greatest loss with its most favorable neighbor. Step 3. Repeat Step 1 and 2 until a tour is formed.

19. Loss or opportunity cost computation -For each city, there are 4 categories that each city falls into.

20. Case A k j i

21. Case B k j i k’ -(j,k) and (j,k’) are the cheapest and the next cheapest arcs emanating from j such that no subtour is formed. -The loss for city j can be computed from

22. Case C i j k I’ -(i,j) and (i’,j) are the cheapest and the next cheapest arcs into city j such that no subtour is formed. -The loss for city j can be computed from

23. Case D i k j k’ I’ -(i,j) and (i’,j) are the cheapest and the next cheapest arcs into city j such that no subtour is formed. -(j,k) and (j,k’) are the cheapest and the next cheapest arcs emanating from j such that no subtour is formed. -The loss for city j can be computed from

24. City 1 2 3 4 5 6 ---------------------------------------------- City1 0 7 4 21 10 15 City2 5 0 11 9 8 14 City3 4 11 0 15 12 15 City4 17 12 20 0 12 10 City5 10 8 12 16 0 5 City6 15 10 15 10 6 0 -----------------------------------------------

25. Depot Customer ปัญหาการจัดเส้นทางของยานพาหนะ Vehicle Routing Problem-VRP • One depot node (origin and destination). • Multiple demand nodes and multiple capacitated vehicles. • Objective: Minimize the total cost (travel time + cost of vehicles + labor) or sometimes minimize the number of vehicles.

26. Depot Customer ปัญหาการจัดเส้นทางของยานพาหนะ Vehicle Routing Problem-VRP General problem To find the set of vehicle routes such that 1. Each route starts and ends at the depot. 2. Each customer node is visited by one vehicle. 3. The total demand of customers visited by a vehicle does not exceed the vehicle capacity.

27. Minimize z = (1) (2) (3) (4) (5) (6) (7)

28. Heuristics for VRP Constructive methods -Construct routes sequentially or in parallel. -Clarke-Wright Savings Heuristic Two-phase methods Phase 1-Cluster the customers by assigning them to vehicles. Phase 2-Route the clusters. -Sweep Heuristic, Christofides’ Heuristic

29. Return the vehicle to the depot if the next demand will violate the vehicle capacity. Start at the Depot and move along the tour meeting demands by one vehicle. Repeat the TSP heuristic separately on all routes to gain any possible savings. Stop. Yes Is there any node not assigned to a vehicle ? VRP Route-First-Cluster-Second Heuristic Use any TSP heuristic to construct a tour through all nodes, including the depot. capacity=100 20 30 35 25 40 10 No 50 30

30. Return the vehicle to the depot if the next demand will violate the vehicle capacity. Start at the Depot and move along the tour meeting demands by one vehicle. Repeat the TSP heuristic separately on all routes to gain any possible savings. Stop. Yes Is there any node not assigned to a vehicle ? VRP Route-First-Cluster-Second Heuristic Use any TSP heuristic to construct a tour through all nodes, including the depot. capacity=100 20 30 2 1 35 25 3 0 8 40 4 5 10 No 7 50 6 30

31. VRP Sweep Heuristic Arbitrarily choose a starting demand node. capacity=100 30 30 Sweep in a clockwise manner from that node, summing the demand that is passed until the next demand to be added would exceed vehicle capacity. 40 20 Is there any node not added in a group? Yes 30 40 No 20 Do the TSP on each cluster, with the depot as the star and end point. 30 Stop.

32. s = 2 s = 3 s = 3 For each possible pair of nodes, calculate the distance saved if the pair is combined onto the same route (eliminate a vehicle). Combine the pair that results in the largest distance saved. s = 5 s = 4 Yes s = 2 s = 7 s = 3 VRP Clark-Wright Saving Heuristic Begin by using one vehicle for each demand node. capacity=50 1 4 5 7 2 Can remaining pairs be combined without violating the vehicle capacity? 3 8 No 6 Stop

33. Clark-Wright Saving Heuristic 8 1 2 1 2 10 14 14 10 0 0 Savings = 10 + 14 – 8 = 16 Total distance = (2x10) + (2x14) = 48 Total distance = 10 + 8 + 14 = 32

34. Clark-Wright Saving Heuristic • 4-7demand = 30+30 = 60 • 0—4—7—0 • distance = 50+10+90 =150 • 1-3 demand = 46+33 = 79 • 0—1—3—0 • distance = 20+10+51 = 81 • 2-5 demand = 55+24 = 79 • 0—2—5—0 • distance = 57+30+10 = 97 • demand= 75 • 0—6—0 distance = 15x2 = 30 • Total distance = 150+81+97+30=358 Savings 1-3 1-4 1-5 1-7 2-5 3-4 3-5 3-7 4-5 4-7 5-7 = 20 + 51 - 10 = 61 = 20 + 50 – 55 = 15 = 20 + 10 – 25 = 5 = 20 + 90 – 53 = 57 = 57 + 10 – 30 = 37 = 51 + 50 – 50 = 51 = 51 + 10 – 11 = 50 = 51 + 90 – 38 = 103 = 50 + 10 – 50 = 10 = 50 + 90 – 10 = 130 = 10 + 90 – 90 = 10

35. Savings algorithm [Clark & Wright, 1964] Step 1. Calculate the savings for all pairs of customers i and j. Step 2. Order the savings in descending order. Step 3. Starting at the top of the list, do the following. Sequential version Step 4. Find the first feasible link in the list which can be used to extend one of the two ends of the currently constructed route. Step 5. If the route cannot be expanded further, terminate the route. Choose the first feasible link in the list to start a new route. Step 6. Repeat Steps 4 and 5 until no more links can be chosen.

36. The two-phase method of Christofides, Mingozzi & Toth [1979] Phase I Step 1. (Sequential trial) Choose an unrouted customer to be a seed. Choose a vehicle k to allocate to the emerging route. Step 2. Enter unrouted customers into emerging cluster, in increasing order of some insertion cost relative to the seed of the cluster, until the capacity limit of vehicle k is reached. If all customers are clustered, or all vehicles used, go to Step 3, else repeat from Step 1. Step 3. (Parallel trial) Using the seeds chosen in the sequential trial, free all customers from their clusters.

37. Step 4. For every free customer, compute its insertion cost into a feasible cluster relative to the seed of the cluster. Consider all clusters and keep the best insertion for the customer. Step 5. Of the free customers, allocate the one with minimum insertion cost to its corresponding cluster. Step 6. Repeat Step 4 for any free customer whose previously nest insertion is no longer feasible, and continue with Step 5 until no further feasible insertions are possible.

38. Phase II Step 7. For both the above two clusterings formed sequentially and in parallel, solve the TSP for each cluster. Keep the best of the two as the VRP solution.

39. Methodology Neighborhood Search • Improvethe solution by performing local search methods. • Usually obtain a local optimal solution. • Types of local search for VRP • Search a better solution (a shorter distance) in one route (2-opt, 3-opt, etc.). • Search a better solution in two or more routes (exchange heuristic, VLSN, etc.).

40. One Move 2 3 2 3 4 4 1 1 1 1 5 5 Depot Depot Depot Depot Swap 2 2 4 4 3 3 5 5 Methodology Neighborhood Search

41. Methodology Neighborhood Search 2- Opt 2 2 3 3 4 4 1 1 5 5 Depot Depot

42. Very Large Scale Neighborhood (VLSN) Search Cyclic Exchange 2 6 10 2 3 1 3 1 9 9 4 4 10 8 5 5 7 8 7 6

43. Very Large Scale Neighborhood (VLSN) Search Path Exchange 2 10 2 3 1 3 1 9 9 4 4 10 8 8 5 5 7 6 7 6

44. กรณีศึกษา บริษัท วารีเทพ จำกัด สาขาอำเภอคำเขื่อนแก้ว จังหวัดยโสธร ผลิตน้ำแข็งยูนิตขายในอำเภอคำเขื่อนแก้ว อำเภอมหาชนะชัย และบริเวณใกล้เคียง

45. รถบรรทุก 4 ล้อ 6 คัน รถบรรทุก 6ล้อ 1 คัน รถขนส่งน้ำแข็ง

46. ส่งสินค้าที่ ต.ลุมพุก ต.แคนน้อย, ต.บากเรือ, ต.โนนทราย, ต.บึงแก มีลูกค้า รวม 72ราย สายการขนส่งรถเบอร์ 3

47. ใช้วิธีหาค่าประหยัด ในการจัดเส้นทางการขนส่ง • ค่าน้ำมัน 1.45บาทต่อกิโลเมตร • ประหยัดค่าน้ำมันได้ =1.45 x 47.175 = 68.40บาทต่อวัน = 68.40 x 30 =2,052บาทต่อเดือน = 2,052 x 12 = 24,624บาทต่อปี

48. Problem Distance Matrix ----------------------------------------------------------------------- Warehouse and Suppliers 0 1 2 3 4 5 6 7 8 9 10 ----------------------------------------------------------------------- warehouse 0 7 4 21 10 15 7 7 9 10 8 Supplier1 7 0 11 24 8 14 4 14 5 7 7 Supplier2 4 11 0 20 12 15 10 3 13 12 10 Supplier3 21 24 20 0 16 10 19 18 20 17 16 Supplier4 10 8 12 16 0 6 4 13 4 1 2 Supplier5 15 14 15 10 6 0 10 14 10 7 7 Supplier6 7 4 10 19 4 10 0 12 2 3 3 Supplier7 7 14 3 18 13 14 12 0 14 13 10 Supplier8 9 5 13 20 4 10 2 14 0 2 5 Supplier9 10 7 12 17 1 7 3 13 2 0 3 Supplier10 8 7 10 16 2 7 3 10 5 3 0 ----------------------------------------------------------------------- -------------------- Demand (boxes) -------------------- Supplier 1 : 20 Supplier 2 : 10 Supplier 3 : 50 Supplier 4 : 25 Supplier 5 : 15 Supplier 6 : 35 Supplier 7 : 40 Supplier 8 : 20 Supplier 9 : 25 Supplier 10 : 30 --------------------

49. Problem • A beer distributor has received orders from seven customers for delivery the next day. The number of cases required by each customer and travel times between each pair of customer are as follows Customer 1 2 3 4 5 6 7 Cases 46 55 33 30 24 75 30 Assume a delivery truck has unlimited capacity. Use the Nearest Neighbor Heuristic and the Arbitrary Insertion Heuristic to construct vehicle routes.

50. Problem 2. A beer distributor has received orders from seven customers for delivery the next day. The number of cases required by each customer and travel times between each pair of customer are as follows Customer 1 2 3 4 5 6 7 Cases 46 55 33 30 24 75 30 Delivery trucks have capacity for 80 cases. Use the Clarke-Wright Savings Heuristic to construct vehicle routes.